RD Sharma Solutions Class 8 Linear Equation In One Variable Exercise 9.3

RD Sharma Solutions Class 8 Chapter 9 Exercise 9.3

RD Sharma Class 8 Solutions Chapter 9 Ex 9.3 PDF Free Download

Solve each of the following equations and also verify your solutions:

\(Q1 \frac{2x – 3}{3x + 2} = -\frac{2}{3}\)

Sol:

\(\frac{2x – 3}{3x + 2} = -\frac{2}{3}\)

=> 6x – 9 = -6x – 4

=> 6x + 6x = 9 – 4

=>     12x = 5

=> x =  \(\frac{5}{12}\)

Verification

\(L.H.S = \frac{2(\frac{5}{12}) – 3}{3(\frac{5}{12}) + 2}\)

\(= \frac{\frac{5}{6} – 3}{\frac{5}{4} + 2}\)

\(= \frac{\frac{-13}{6})}{\frac{13}{4}}\)

\(= \frac{-4}{6}\)

\(= \frac{-2}{3}\)

\(R.H.S = \frac{-2}{3}\)

Hence, L.H.S = R.H.S

\(Q2. \frac{2 – y}{y + 7} = \frac{3}{5}\)

Sol:

\(\frac{2 – y}{y + 7} = \frac{3}{5}\)

=> 10 – 5y = 3y + 21

=> 3y + 5y = 10 – 21

=>    8y = – 11

=>      y = \(\frac{-11}{8}\)

Verification

\(L.H.S = \frac{2 – (\frac{-11}{8})}{(\frac{-11}{8}) + 7}\)

\(= \frac{16 + 11}{-11 + 56}\)

\(= \frac{27}{45}\)

\(= \frac{3}{5}\)

\(R.H.S = \frac{3}{5}\)

Hence, L.H.S = R.H.S

\(Q3 \frac{5x – 7}{3x} = 2\)

Sol:

\(\frac{5x – 7}{3x} = 2\)

=> 6x = 5x – 7

=> 6x – 5x = 7

=>     x = -7

Verification

L.H.S = \(\frac{5(-7) – 7}{3(-7)}\)

= \(\frac{-35 – 7}{-21}\)

= \(\frac{-42}{-21}\)

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q4 \(\frac{3x + 5}{2x + 7}\) = 4

Sol:

\(\frac{3x + 5}{2x + 7}\) = 4

=> 3x + 5 = 8x + 28

=> 8x – 3x = 5 – 28

=> 5x = – 23

=>     x = \(\frac{-23}{5}\)

Verification

L.H.S = \(\frac{3(\frac{-23}{5}) + 5}{2(\frac{-23}{5}) + 7}\)

= \(\frac{-69 + 25}{-46 + 35}\)

= \(\frac{-44}{-11}\)

= 4

R.H.S = 4

Hence, L.H.S = R.H.S

Q5 \(\frac{2y + 5}{y + 4}\) = 1

Sol:

\(\frac{2y + 5}{y + 4}\)

=> 2y + 5 = y + 4

=> 2y – y = 4 – 5

=> y = -1

Verification

L.H.S = \(\frac{2(-1) + 5}{(-1) + 4}\)

= \(\frac{-2 + 5}{3}\)

= \(\frac{3}{3}\)

= 1

R.H.S = 1

Hence, L.H.S = R.H.S

Q6 \(\frac{2x + 1}{3x – 2}\) = \(\frac{5}{9}\)

Sol:

\(\frac{2x + 1}{3x – 2}\) = \(\frac{5}{9}\)

=> 18x + 9 = 15x – 10

=> 18x – 15x = -10 – 9

=>              3x = -19

=>                x = \(\frac{-19}{3}\)

Verification

L.H.S = \(\frac{2(\frac{-19}{3}) + 1}{3(\frac{-19}{3}) – 2}\)

= \(\frac{-38 + 3}{-57 – 6}\)

= \(\frac{-35}{-63}\)

= \(\frac{5}{9}\)

R.H.S = \(\frac{5}{9}\)

Hence, L.H.S = R.H.S

Q7 \(\frac{1 – 9y}{19 – 3y}\) = \(\frac{5}{8}\)

Sol:

\(\frac{1 – 9y}{19 – 3y}\) = \(\frac{5}{8}\)

=> 8 – 72y = 95 – 15y

=> 72y – 15y = 8 – 95

=>        57y = -87

=>            y = \(\frac{-87}{57}\)

=>             y = \(\frac{-29}{19}\)

Verification

L.H.S = \(\frac{1 – 9(\frac{-29}{19})}{19 – 3(\frac{-29}{19})}\)

= \(\frac{19 + 261}{361 + 87}\)

= \(\frac{280}{448}\)

= \(\frac{5}{8}\)

R.H.S = \(\frac{5}{8}\)

Hence, L.H.S = R.H.S

Q8 \(\frac{2x}{3x + 1}\) = -3

Sol:

\(\frac{2x}{3x + 1}\) = -3

=> 2x = -9x – 3

=>  2x + 9x = -3

=> 11x = -3

=>      x = \(\frac{-3}{11}\)

Verification

L.H.S = \(\frac{2(\frac{-3}{11})}{3(\frac{-3}{11}) + 1}\)

= \(\frac{-6}{-9 + 11}\)

= \(\frac{-6}{2}\)

= -3

R.H.S = -3

Hence, L.H.S = R.H.S

Q9 \(\frac{y – (7 – 8y) }{9y – (3 + 4y)}\) = \(\frac{2}{3}\)

Sol:

\(\frac{y – (7 – 8y) }{9y – (3 + 4y)}\) = \(\frac{2}{3}\)

=> \(\frac{9y – 7}{5y – 3}\) = \(\frac{2}{3}\)

=> 27y – 21 = 10y – 6

=> 27y – 10y = 21 – 6

=>  17y = 15

=>      y = \(\frac{15}{17}\)

Verification

L.H.S = \(\frac{9(\frac{15}{17}) – 7}{5(\frac{15}{17}) – 3}\)

=  \(\frac{135 – 119}{75 – 51}\)

= \(\frac{16}{24}\)

= \(\frac{2}{3}\)

R.H.S = \(\frac{2}{3}\)

Hence, L.H.S = R.H.S

Q10 \(\frac{6}{2x – 3(3 – 4x)}\) = \(\frac{2}{3}\)

Sol:

\(\frac{6}{2x – 3(3 – 4x)}\) = \(\frac{2}{3}\)

=> \(\frac{6}{6x – 3}\) = \(\frac{2}{3}\)

=> 12x – 6 = 18

=> 12x = 18 + 6

=> 12x = 24

=>      x = \(\frac{24}{12}\)

=>      x = 2

Verification

L.H.S = \(\frac{6}{2(2) – 3(3 – 4(2))}\)

= \(\frac{6}{9}\)

= \(\frac{2}{3}\)

R.H.S = \(\frac{2}{3}\)

Hence, L.H.S = R.H.S

Q11 \(\frac{3}{2x}\)\(\frac{3}{2x}\) = \(\frac{1}{12}\)

Sol:

\(\frac{3}{2x}\)\(\frac{3}{2x}\) = \(\frac{1}{12}\)

=> \(\frac{4 – 9}{6x}\) = \(\frac{1}{12}\)

=> \(\frac{-5}{x}\) = \(\frac{1}{2}\)

=>         x = -10

Verification

L.H. S = \(\frac{3}{2(-10)}\)\(\frac{3}{2(10)}\)

= \(\frac{2}{-30}\)\(\frac{3}{-30}\)

= \(\frac{4 – 9}{-60}\)

= \(\frac{-5}{-60}\)

= \(\frac{1}{12}\)

R.H.S = \(\frac{1}{12}\)

Hence, L.H.S = R.H.S

Q12 \(\frac{3x + 5}{4x + 2}\) = \(\frac{3x + 4}{4x + 7}\)

Sol:

\(\frac{3x + 5}{4x + 2}\) = \(\frac{3x + 4}{4x + 7}\)

=> \(12x^{2} + 20x + 21x + 35 = 12x^{2} + 16x + 6x + 8\)

=> \(12x^{2} – 12x^{2} + 41x – 22x = 8 – 35\)

=> 19x = -27

=>     x = \(\frac{-27}{19}\)

Verification

L.H.S = \(\frac{3(\frac{-27}{19}) + 5}{4(\frac{-27}{19}) + 2}\)

= \(\frac{-81 + 95}{-108 + 38}\)

= \(\frac{14}{-70}\)

= \(\frac{-1}{5}\)

R.H.S = \(\frac{3(\frac{-27}{19}) + 4}{4(\frac{-27}{19}) + 7}\)

= \(\frac{-81 + 76}{-108 + 133}\)

= \(\frac{-5}{25}\)

= \(\frac{-1}{5}\)

Hence, L.H.S = R.H.S

Q13 \(\frac{7x – 2}{5x – 1}\) = \(\frac{7x + }{5x + 4}\)

Sol:

\(\frac{7x – 2}{5x – 1}\) = \(\frac{7x + }{5x + 4}\)

=> \(35x^{2} + 28x – 10x – 8 = 35x^{2} + 15x – 7x – 3\)

=> \(35x^{2} – 35x^{2} + 18x – 8x = 8 – 3\)

=> 10x = 5

=>    x = \(\frac{5}{10}\)

= \(\frac{1}{2}\)

Verification

L.H.S = \(\frac{7(\frac{1}{2}) – 2}{5(\frac{1}{2}) – 1}\)

= \(\frac{7 – 4}{5 – 2}\)

= \(\frac{3}{3}\)

= 1

R.H.S = \(\frac{7(\frac{1}{2}) + 3}{5(\frac{1}{2})  + 4}\)

= \(\frac{7 + 6}{5 + 8}\)

= \(\frac{3}{3}\)

= 1

Hence, L.H.S= R.H.S

Q14 \((\frac{x + 1}{x + 2})^{2}\) = \(\frac{x + 2}{x + 4}\)

Sol:

\((\frac{x + 1}{x + 2})^{2}\) = \(\frac{x + 2}{x + 4}\)

=> \(\frac{x^{2} + 2x + 1}{x^{2} + 4x + 4}\) = \(\frac{x + 2}{x + 4}\)

=> \(x^{3} + 2x^{2} + x + 4x^{2} + 8x + 4 = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8\)

=> \(x^{3} – x^{3} + 6x^{2} – 6x^{2} + 9x – 12x = 8 – 4\)

=> -3x = 4

=>     x = \(\frac{-4}{3}\)

Verification

L.H.S = \((\frac{\frac{-4}{3} + 1}{\frac{-4}{3} + 2})^{2}\)

= \((\frac{-4 + 3}{-4 + 6})^{2}\)

= \(\frac{1}{4}\)

R.H.S = \(\frac{\frac{-4}{3} + 1}{\frac{-4}{3} + 2}\)

= \(\frac{-4 + 6}{-4 + 12}\)

= \(\frac{2}{8}\)

= \(\frac{1}{4}\)

Hence, L.H.S = R.H.S

Q15 \((\frac{x + 1}{x – 4})^{2}\) = \(\frac{x + 8}{x – 2}\)

Sol:

\((\frac{x + 1}{x – 4})^{2}\) = \(\frac{x + 8}{x – 2}\)

=> \(\frac{x^{2} + 2x + 1}{x^{2} –  8x + 16}\) = \(\frac{x + 8}{x – 2}\)

=> \(x^{3} + 2x^{2} + x – 2x^{2} – 4x – 2 = x^{3} – 8x^{2} + 16x + 8x^{2} – 64x + 128\)

=> \(x^{3} – x^{3} – 3x + 48x = 128 + 2\)

=> 45x = 130

=>     x = \(\frac{130}{45}\)

= \(\frac{26}{9}\)

Verification

L.H.S = \((\frac{\frac{26}{9} + 1}{\frac{26}{9} – 4})^{2}\)

= \((\frac{26 + 9}{26 – 36})^{2}\)

= \(\frac{1225}{100}\)

= \(\frac{49}{4}\)

R.H.S = \(\frac{\frac{26}{9} + 8}{\frac{26}{9} – 2}\)

= \(\frac{26 + 72}{26 – 18}\)

= \(\frac{98}{8}\)

= \(\frac{49}{4}\)

Hence, L.H.S = R.H.S

Q16 \(\frac{9x – 7}{3x + 5}\) = \(\frac{3x – 4}{x + 6}\)

Sol:

\(\frac{9x – 7}{3x + 5}\) = \(\frac{3x – 4}{x + 6}\)

=> \(9x^{2} – 7x +  54x – 42 = 9x^{2} – 12x + 15x – 20\)

=> \(9x^{2} – 9x^{2} + 47x – 3x = –20 + 42\)

=> 44x = 22

=>     x = \(\frac{22}{44}\)

= \(\frac{1}{2}\)

Verification

L.H.S = \(\frac{9(\frac{1}{2}) – 7}{3\frac{1}{2} + 5}\)

= \(\frac{9 – 14}{3 + 10}\)

= \(\frac{-5}{3}\)

R.H.S = \(\frac{3(\frac{1}{2}) – 4}{\frac{1}{2} + 6}\)

= \(\frac{3 – 8}{1 + 12}\)

= \(\frac{-5}{13}\)

Hence, L.H.S = R.H.S

Q17 \(\frac{x + 2}{x + 5}\) = \(\frac{x}{x + 6}\)

Sol:

\(\frac{x + 2}{x + 5}\) = \(\frac{x}{x + 6}\)

=> \(x^{2} + 2x +  6x + 12 = x^{2} + 5x\)

=> \(x^{2} – x^{2} + 8x – 5x = -12\)

=>   3x = – 12

=>      x = \(\frac{-12}{3}\)

=>       x = -4

Verification

L.H.S = \(\frac{-4 + 2}{-4 + 5}\)

= -2

R.H.S = \(\frac{-4}{-4 + 6}\)

= -2

Hence, L.H.S = R.H.S

Q18 \(\frac{2x – (7 – 5x)}{9x – (3 + 4x) }\) = \(\frac{7}{6}\)

Sol:

\(\frac{2x – (7 – 5x)}{9x – (3 + 4x) }\) = \(\frac{7}{6}\)

=> \(\frac{7x – 7}{5x – 3}\) = \(\frac{7}{6}\)

=> 42x – 42 = 35x – 21

=> 42x – 35x = 42 – 21

=> 7x = 21

=>    x = \(\frac{21}{7}\)

= 3

Verification

L.H.S = \(\frac{2(3) – (7 – 5(3))}{9(3) – (3 + 4(3)) }\)

= \(\frac{6 – (7 – 15)}{27 – (3 + 12)}\)

= \(\frac{6 + 8}{27 – 15}\)

= \(\frac{14}{12}\)

= \(\frac{7}{6}\)

R.H.S = \(\frac{7}{6}\)

Hence, L.H.S = R.H.S

Q19 \(\frac{15(2 – x) – 5(x + 6)}{1 – 3x}\) = 10

Sol:

\(\frac{15(2 – x) – 5(x + 6)}{1 – 3x}\) = 10

=> \(\frac{-20x}{1 – 3x}\) = 10

=> 10 – 30x = -20x

=> 30x – 20x = 10

=>        10x = 10

=>            x = 1

Verification

L.H.S = \(\frac{15(2 – 1) – 5(1 + 6)}{1 – 3(1)}\)

= \(\frac{15 – 35}{-2}\)

= \(\frac{-20}{-2}\)

= 10

R.H.S = 10

Hence, L.H.S = R.H.S

Q20  \(\frac{x + 3}{x – 3}\) + \(\frac{x + 2}{x – 2}\) = 2

Sol:

\(\frac{x + 3}{x – 3}\) + \(\frac{x + 2}{x – 2}\) = 2

=> \(\frac{x + 3}{x – 3}\) = 2 – \(\frac{x + 2}{x – 2}\)

=> \(\frac{x + 3}{x – 3}\) = \(\frac{2x – 4 – x – 2}{x – 2}\)

=>\(\frac{x + 3}{x – 3}\) = \(\frac{x – 6}{x – 2}\)

=> \(x^{2} – 2x +  3x – 6 = x^{2} – 3x – 6x + 18\)

=> \(x^{2} – x^{2} + x + 9x = 18 + 6\)

=>             10x = 24

=>                 x = \(\frac{24}{10}\)

=>                  x = \(\frac{12}{5}\)

Verification

L.H.S = \(\frac{\frac{12}{5} + 3}{\frac{12}{5} – 3}\) + \(\frac{\frac{12}{5} + 2}{\frac{12}{5} – 2}\)

= \(\frac{12 + 15}{12 – 15}\) + \(\frac{12 + 10}{12 – 10}\)

= \(\frac{27}{-3}\) + \(\frac{22}{2}\)

= \(\frac{54 – 66}{-6}\)

= \(\frac{-12}{-6}\)

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q21 \(\frac{(x + 2)(2x – 3) – 2x^{2} + 6}{x – 5}\) = 2

Sol:

\(\frac{(x + 2)(2x – 3) – 2x^{2} + 6}{x – 5}\) = 2

=> \(\frac{2x^{2} + x – 6 – 2x^{2} + 6}{x – 5}\) = 2

=> \(\frac{x}{x – 5}\) = 2

=> 2x – 10 = x

=> 2x – x = 10

=>        x = 10

Verification

L.H.S = \(\frac{(10 + 2)(2(10) – 3) – 2(10)^{2} + 6}{10 – 5}\)

= \(\frac{12(17) – 200 + 6}{5}\)

= \(\frac{10}{5}\)

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q22 \(\frac{x^{2} – (x + 1) (x + 2)}{5x + 1}\) = 6

Sol:

\(\frac{x^{2} – (x + 1) (x + 2)}{5x + 1}\) = 6

=> \(\frac{x^{2} – x^{2} – 2x – x – 2}{5x + 1}\) = 6

=> \(\frac{-3x – 2}{5x + 1}\) = 6

=> 30x + 6 = -3x – 2

=> 30x + 3x = -2 – 6

=>     33x = -8

=>         x = \(\frac{-8}{33}\)

Verification

L.H.S = \(\frac{(\frac{-8}{33})^{2} – (\frac{-8}{33} + 1)(\frac{-8}{33} + 2)}{5(\frac{-8}{33}) + 1}\)

= \(\frac{\frac{64}{1089} – \frac{25}{33}(\frac{58}{33})}{\frac{-49}{33} + 1}\)

= \(\frac{\frac{64}{1089} – \frac{1450}{1089}}{\frac{-7}{33}}\)

= \(\frac{\frac{-1386}{1089}}{\frac{-7}{33}}\)

= \(\frac{42}{7}\)

= 0

R.H.S = 0

Hence, L.H.S = R.H.S

Q23 \(\frac{(2x + 3) – (5x – 7)}{6x + 11}\) = \(\frac{-8}{3}\)

Sol:

\(\frac{(2x + 3) – (5x – 7)}{6x + 11}\) = \(\frac{-8}{3}\)

=> \(\frac{-3x + 10}{6x + 11}\) = \(\frac{-8}{3}\)

=> -9x + 30 = -48x – 88

=> 48x – 9x = -88 – 30

=>   39x = -118

=>      x = \(\frac{-118}{39}\)

Verification

L.H.S = \(\frac{-3(\frac{-118}{39}) + 10}{6(\frac{-118}{39}) + 11}\)

= \(\frac{354 + 390}{-708 + 429}\)

= \(\frac{744}{-279}\)

= \(\frac{-8}{3}\)

R.H.S = \(\frac{-8}{3}\)

Hence, L.H.S = R.H.S

Q24 Find the positive value of x for which the given equations is satisfied

(i) \(\frac{x^{2} – 9}{5 + x^{2}} = \frac{-5}{9}\)

Sol:

\(\frac{x^{2} – 9}{5 + x^{2}} = \frac{-5}{9}\)

=> \(9x^{2} – 81 = -25 – 5x^{2}\)

=> \(9x^{2} + 5x^{2} = 81 – 25\)

=> \(14x^{2} = 56\)

=> \(x^{2} = \frac{56}{14}\)

=> \(x^{2} = 4\)

=>           x = 2

Verification

L.H.S = \(\frac{2^{2} – 9}{5 + 2^{2}}\)

= \(\frac{4 – 9}{5 + 4}\)

= \(\frac{-5}{9}\)

R.H.S = \(\frac{-5}{9}\)

Hence, L.H.S = R.H.S

(ii) \(\frac{y^{2} + 4}{3y^{2} + 7} = \frac{1}{2}\)

Sol:

\(\frac{y^{2} + 4}{3y^{2} + 7} = \frac{1}{2}\)

=> \(3y^{2} + 7 = 2y^{2} + 8\)

=> \(3y^{2} – 2y^{2} = 8 – 7\)

=> \(y^{2} = 1\)

=>  y = 1

Verification

L.H.S = \(\frac{1^{2} + 4}{3(1)^{2} + 7}\)

= \(\frac{5}{10}\)

= \(\frac{1}{2}\)

R.H.S = \(\frac{1}{2}\)

Hence, L.H.S = R.H.S